Essential Cell Biology 5th edition

A:12 Answers
which it would normally be stable. Polypeptide chains that
denature when the temperature is raised often aggregate,
and they rarely refold into active proteins when the
temperature is decreased.
ANSWER 4–19 The motor protein in the illustration can
move just as easily to the left as to the right and so will not
move steadily in one direction. However, if just one of the
steps is coupled to ATP hydrolysis (for example, by making
detachment of one foot dependent on binding of ATP and
coupling the reattachment to hydrolysis of the bound ATP),
then the protein will show unidirectional movement that
requires the continued consumption of ATP. Note that, in
principle, it does not matter which step is coupled to ATP
hydrolysis (Figure A4–19).
rate of reaction (µmole/sec)
100
50
0 1 5 10
substrate (mM)
0.07
ATP
ADP
+
P
sec
µmole
0.05
1
rate
0.03
Figure A4−19
ANSWER 4–20 The slower migration of small molecules
through a gel-filtration column occurs because smaller
molecules have access to many more spaces in the porous
beads that are packed
ECB5
into
EA4.19/A4.19
the column than do larger
molecules. However, it is important that the flow rate
through the column is slow enough to give the smaller
molecules sufficient time to diffuse into the spaces inside
the beads. At very rapid flow rates, all molecules will move
rapidly around the beads, so that large and small molecules
will now tend to exit together from the column.
ANSWER 4–21 The α helix in the figure is right-handed,
whereas the coiled-coil is left-handed. The reversal occurs
because of the staggered positions of hydrophobic side
chains in the α helix.
ANSWER 4–22 The atoms at the binding sites of proteins
must be precisely located to fit the molecules that they
bind. Their location in turn requires the precise positioning
of many of the amino acids and their side chains in the core
of the protein, distant from the binding site itself. Thus,
even a small change in this core can disrupt protein function
by altering the conformation at a binding site far away.
ANSWER 4–23
A. When [S] << K M , the term (K M + [S]) approaches K M .
Therefore, the equation is simplified to rate =
V max [S]/K M . Therefore, the rate is proportional to [S].
B. When [S] = K M , the term [S]/(K M + [S]) equals ½.
Therefore, the reaction rate is half of the maximal rate
V max .
C. If [S] >> K M , the term (K M + [S]) approaches [S].
Therefore, [S]/(K M + [S]) equals 1 and the reaction occurs
at its maximal rate V max .
ANSWER 4–24 The substrate concentration is 1 mM.
This value can be obtained by substituting values into the
equation, but it is simpler to note that the desired rate
(50 μmole/sec) is exactly half of the maximum rate, V max ,
where the substrate concentration is typically equal to the
K M . The two plots requested are shown in Figure A4–24.
0.01
0 1 2 3 4 5
1 1
[S] mM
Figure A4−24
A plot of 1/rate versus 1/[S] is a straight line because
rearranging the standard equation yields the equation given
in Question 4–26B.
ANSWER 4–25 If [S] is very much smaller than K M , the
active site of the enzyme is mostly unoccupied. If [S] is very
much greater than K M , the reaction rate is limited by the
enzyme concentration (because most of the catalytic sites
are fully occupied).
ANSWER 4–26
A,B. The data in the boxes have been used to plot the red
curve and red line in Figure A4–26. From the plotted
data, the K M is 1 μM and the V max is 2 μmole/min. Note
that the data are much easier to interpret in the linear
plot, because the curve in (A) approaches, but never
reaches, V max .
C. It is important that only a small quantity of product is
made, because otherwise the rate of reaction would
decrease as the substrate was depleted and product
accumulated. Thus the measured rates would be lower
than they should be.
D. If the K M increases, then the concentration of substrate
needed to give a half-maximal rate is increased. As
more substrate is needed to produce the same rate,
the enzyme-catalyzed reaction has been inhibited by
the phosphorylation. The expected data plots for the
phosphorylated enzyme are the green curve and the
green line in Figure A4–26.
Chapter 5
ANSWER 5–1
A. False. The polarity of a DNA strand commonly refers to
the orientation of its sugar–phosphate backbone, one
end of which contains a phosphate group and the other
a hydroxyl group.